You may use alternative method, hence you should expand the sine of sum such that:

`sin(pi/2 + h) = sin (pi/2)cos h + sin h*cos (pi/2)`

Since `cos (pi/2) = 0` , hence the product `sin h*cos (pi/2) = 0` .

You need to substitute 1 for `sin(pi/2)` such that:

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You may use alternative method, hence you should expand the sine of sum such that:

`sin(pi/2 + h) = sin (pi/2)cos h + sin h*cos (pi/2)`

Since `cos (pi/2) = 0` , hence the product `sin h*cos (pi/2) = 0` .

You need to substitute 1 for `sin(pi/2)` such that:

`sin(pi/2 + h) = cos h`

You need to write the limit in terms of cos h such that:

`lim_(h-gt0) (cos h - 1)/h`

You need to remember the formula of half of angle such that:

`1 - cos h = 2 sin^2 (h/2) =gt cos h - 1 = -2 sin^2 (h/2)`

You need to substitute`-2 sin^2 (h/2)` for `cos h - 1` in limit such that:

`lim_(h-gt0) (-2 sin^2 (h/2))/h = -2lim_(h-gt0) (sin^2 (h/2))/h`

You need to use special limit `lim_(x-gt0) (sin (x))/x = 1` such that:

`-2 lim_(h-gt0) (sin^2 (h/2))/h = -2 lim_(h-gt0) (sin^2 (h/2))/(2*(h/2)) = (-2/2)lim_(h-gt0) (sin(h/2))/(h/2)*lim_(h-gt0) (sin(h/2))`

`-2 lim_(h-gt0) (sin^2 (h/2))/h = -1*0=0`

**Hence, evaluating the limit of the fraction yields `lim_(h-gt0) (sin(pi/2 + h)-sin(pi/2))/h = 0` .**

The limit `lim_(h->0) (sin(pi/2+h)-sin (pi/2))/h` has to be determined.

sin (pi/2 + h) = cos h and sin pi/2 = 1

`lim_(h->0) (sin(pi/2+h)-sin (pi/2))/h`

=> `lim_(h->0) (cos h - 1)/h`

substituting h = 0 gives the form 0/0 which is indeterminate. The l'Hopital's rule can be used and the numerator and denominator substituted by their derivative

=> `lim_(h->0) -sin h`

substituting h = 0

=> 0

**The required value of** `lim_(h->0) (sin(pi/2+h)-sin pi/2)/h = 0`